\newproblem{lay:7_3_3}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.3.3}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $Q(\mathbf{x})=5x_1^2+6x_2^2+7x_3^2+4x_1x_2-4x_2x_3=9y_1^2+6y_2^2+3y_3^2$ (see Exercise 7.3.1).
	\begin{enumerate}[a.]
		\item Find the maximum value of $Q(\mathbf{x})$ subject to the constraint $\mathbf{x}^T\mathbf{x}=1$.
		\item Find a unit vector $\mathbf{u}$ where this maximum is attained.
		\item Find the maximum of $Q(\mathbf{x})$ subject to the constraints $\mathbf{x}^T\mathbf{x}=1$ and $\mathbf{x}^T\mathbf{u}=0$
	\end{enumerate}
}{
   % Solution
	\begin{enumerate}[a.]
		\item The maximum value of $Q(\mathbf{x})$ subject to the constraint $\mathbf{x}^T\mathbf{x}=1$ is given by the maximum eigenvalue (see Exercise 7.3.1), which is 9.
		\item The unit vector $\mathbf{u}$ where this maximum is attained is given by the eigenvector associated to this eigenvalue (see Exercise 7.3.1), that is,
					$\mathbf{u}=(-\frac{1}{3},-\frac{2}{3},\frac{2}{3})$.
		\item The maximum of $Q(\mathbf{x})$ subject to the constraints $\mathbf{x}^T\mathbf{x}=1$ and $\mathbf{x}^T\mathbf{u}=0$ is the second eigenvalue of $A$, that is, 6.
		      This value is attained for its corresponding eigenvector, $\mathbf{x}=(\frac{2}{3},\frac{1}{3},\frac{2}{3})$
	\end{enumerate}
}
\useproblem{lay:7_3_3}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

